Moving boundary diffusion equation (transformation of coordinates)

[babagoslow]

I'm trying to implement a numerical code for the diffusion equation with moving boundaries. I have no problems with the numerical implementation, but with the transformation of coordinates. In spherical coordinates, the diffusion equation is
$\frac{\partial c}{\partial t} = D \left(\frac{2}{r}\frac{\partial c}{\partial r} + \frac{\partial^2 c}{\partial r^2} \right)$.

After scaling these equations with $\phi = c/C_0, \tau = Dt/R_0^2, x = (r-R_0)/R_0$ I get

$\frac{\partial\phi}{\partial \tau} = \frac{2}{1+x} \frac{\partial\phi}{\partial x} + \frac{\partial^2 \phi}{\partial x^2}$

Now transform the $x$ coordinate to the Landau (not that Landau) coordinate $\eta(x,\tau)$ defined as

$\eta(x,\tau) = \frac{x - X(\tau)}{X_\infty - X(\tau)}$.

The idea of the transformation is to move the boundary $X(\tau)$, which is variable only in time $\tau$ and keep the length scale $X_\infty$ fixed. This is particularly useful, for instance, in problems involving a gas bubble losing volume by diffusion, where the bubble's radius is variable in time.

There's a paper out there (Fischer, Zinovik and Poulikakos 2009) which suggests that the final result should be

$\frac{\partial \phi}{\partial \tau} = \left[ \frac{2}{(X_\infty - X(\tau))(1+\eta (X_\infty - X(\tau))+ X)} + \frac{1-\eta}{X_\infty - X(\tau)}\frac{dX}{d\tau} \right] \frac{\partial\phi}{\partial\eta} + \frac{1}{(X_\infty - X(\tau))^2} \frac{\partial^2 \phi}{\partial \eta^2}.$

However, I have some trouble reproducing this result. Can someone guide me through the change of coordinates? I have problems particularly with the second partial derivative.

 

[pasmith]

I'm trying to implement a numerical code for the diffusion equation with moving boundaries. I have no problems with the numerical implementation, but with the transformation of coordinates. In spherical coordinates, the diffusion equation is
$\frac{\partial c}{\partial t} = D \left(\frac{2}{r}\frac{\partial c}{\partial r} + \frac{\partial^2 c}{\partial r^2} \right)$.

After scaling these equations with $\phi = c/C_0, \tau = Dt/R_0^2, x = (r-R_0)/R_0$ I get

$\frac{\partial\phi}{\partial \tau} = \frac{2}{1+x} \frac{\partial\phi}{\partial x} + \frac{\partial^2 \phi}{\partial x^2}$

Now transform the $x$ coordinate to the Landau (not that Landau) coordinate $\eta(x,\tau)$ defined as

$\eta(x,\tau) = \frac{x - X(\tau)}{X_\infty - X(\tau)}$.

The idea of the transformation is to move the boundary $X(\tau)$, which is variable only in time $\tau$ and keep the length scale $X_\infty$ fixed. This is particularly useful, for instance, in problems involving a gas bubble losing volume by diffusion, where the bubble's radius is variable in time.

There's a paper out there (Fischer, Zinovik and Poulikakos 2009) which suggests that the final result should be

$\frac{\partial \phi}{\partial \tau} = \left[ \frac{2}{(X_\infty - X(\tau))(1+\eta (X_\infty - X(\tau))+ X)} + \frac{1-\eta}{X_\infty - X(\tau)}\frac{dX}{d\tau} \right] \frac{\partial\phi}{\partial\eta} + \frac{1}{(X_\infty - X(\tau))^2} \frac{\partial^2 \phi}{\partial \eta^2}.$

However, I have some trouble reproducing this result. Can someone guide me through the change of coordinates? I have problems particularly with the second partial derivative.

It helps, to avoid confusion, to change to two new independent variables, so let's take $(x,\tau) \to (\eta, \sigma)$ with $\sigma(x,\tau) = \tau$.

The chain rule gives
$$\frac{\partial \phi}{\partial x} = \frac{\partial \phi}{\partial \eta} \frac{\partial \eta}{\partial x} + \frac{\partial \phi}{\partial \sigma} \frac{\partial \sigma}{\partial x}<br /> = \frac{1}{X_\infty - X(\sigma)} \frac{\partial \phi}{\partial \eta} \\<br /> \frac{\partial \phi}{\partial \tau} = \frac{\partial \phi}{\partial \eta} \frac{\partial \eta}{\partial \tau} + \frac{\partial \phi}{\partial \sigma} \frac{\partial \sigma}{\partial \tau}<br /> = -\frac{(1- \eta) X'(\sigma)}{X_\infty - X(\sigma)} \frac{\partial \phi}{\partial \eta} + \frac{\partial \phi}{\partial \sigma}$$
since
$$\frac{\partial \eta}{\partial \tau} = \frac{\partial}{\partial \tau} \left(\frac{x - X(\tau)}{X_\infty - X(\tau)}\right) = <br /> \frac{1}{(X_\infty - X)^2} ((X_\infty - X)(-X') - (x - X)(-X')) \\<br /> = \frac{-X'(\sigma)}{X_\infty - X(\sigma)} \frac{X_\infty - x}{X_\infty - X(\sigma)}<br /> = \frac{-X'(\sigma)}{X_\infty - X(\sigma)}(1-\eta)$$

For the second derivative,
$$\frac{\partial^2 \phi}{\partial x^2} = \frac{\partial}{\partial x} \frac{\partial \phi}{\partial x}<br /> = \frac{\partial}{\partial x} \left( \frac{1}{X_\infty - X(\sigma)} \frac{\partial \phi}{\partial \eta} \right) \\<br /> = \frac{\partial \eta}{\partial x} \frac{\partial}{\partial \eta} \left( \frac{1}{X_\infty - X(\sigma)} \frac{\partial \phi}{\partial \eta} \right) <br /> + \frac{\partial \sigma}{\partial x} \frac{\partial}{\partial \sigma} \left( \frac{1}{X_\infty - X(\sigma)} \frac{\partial \phi}{\partial \eta} \right) \\<br /> = \frac{1}{(X_\infty - X(\sigma))^2} \frac{\partial^2 \phi}{\partial \eta^2}$$

Substitution into the original equation and rearrangement will yield the given result.

 

[babagoslow]

Hi Pasmith, thanks, I understand it now!